Question

What is the total distance covered by the body at the end of the journey as shown in the graph.

75 m

67.5 m

100 m

none of the above

​

Answer 1

Given:

• A velocity time graph is given .

To Find:

• The total displacement covered by the body at the end of the journey.

Concept Used:

• As we know that the area of ( v - t ) graph gives us displacement , so we will be finding the area of the graph .

Answer:

(For graph refer to the attachment:)

We can here find the area of the graph in three steps ,

• Firstly finding area of ∆ ABE .
• Secondly finding area of rectangle BCEF.
• Thirdly finding area of ∆ DFC .

Note:On finding area power of dimension will be 1 .

$\red{\bf{\leadsto Step\:1}}$ Finding area of ∆ ABE

Here ,

• Height = 15m/s.
• Base = 2s

$\sf{\implies ar(\Delta ABE )= \dfrac{1}{2}\times (base)\times (height) }$

$\sf{\implies ar(\Delta ABE )= \dfrac{1}{2}\times 2 s\times 15m/s}$

$\boxed{\red{\bf{\longmapsto ar(\triangle ABE )= 15m}}}$

$\rule{200}5$

$\red{\bf{\leadsto Step\:2}}$ Finding area of rectangle BCEF

Here ,

• Breadth = (5-2)s = 3s.
• Length = 15m/s.

$\sf{\implies ar(BCEF) = (length)\times (breadth)}$

$\sf{\implies ar(BCEF) = 3s \times 15m/s }$

$\boxed{\red{\bf{\longmapsto ar( BCEF)= 45m}}}$

$\rule{200}5$

$\red{\bf{\leadsto Step\:3}}$ Finding area of ∆ DFC.

Here ,

• Height = 15m/s.
• Base = (6-5)s = 1s.

$\sf{\implies ar(\Delta DFC) = \dfrac{1}{2}\times (base)\times (height) }$

$\sf{\implies ar(\Delta DFC)= \dfrac{1}{2}\times 1 s\times 15m/s}$

$\boxed{\red{\bf{\longmapsto ar(\triangle DFC )= 7.5m}}}$

$\rule{200}5$

Hence total area = displacement

= ar(∆DFC) + ar(BCEF) + ar(∆ABE).

= 15m + 45m + 7.5m

= 67.5m.

Hence the total displacement is 67.5m .